Difference between revisions of "2018 AIME II Problems/Problem 12"
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− | Either <math>PA=PC</math> or <math>PD=PB</math>. Let <math>PD=PB=s</math>. Applying Stewart's Theorem on <math>\triangle ABD</math> and <math>\triangle BCD</math>, dividing by <math>2s</math> and rearranging, <cmath>\tag{1}CP^2+s^2=148</cmath> <cmath>\tag{2}AP^2+s^2=180</cmath> Applying Stewart on <math>\triangle CAB</math> and <math>\triangle CAD</math>, <cmath>\tag{3} 5CP^2=3AP^2</cmath> Substituting equations 1 and 2 into 3 and rearranging, <math>s=\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}</math> . By Law of Cosines on <math>\triangle APB</math>, <math>\cos(\angle APB)=\frac{4\sqrt{65}}{65}</math> | + | Either <math>PA=PC</math> or <math>PD=PB</math>. Let <math>PD=PB=s</math>. Applying Stewart's Theorem on <math>\triangle ABD</math> and <math>\triangle BCD</math>, dividing by <math>2s</math> and rearranging, <cmath>\tag{1}CP^2+s^2=148</cmath> <cmath>\tag{2}AP^2+s^2=180</cmath> Applying Stewart on <math>\triangle CAB</math> and <math>\triangle CAD</math>, <cmath>\tag{3} 5CP^2=3AP^2</cmath> Substituting equations 1 and 2 into 3 and rearranging, <math>s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}</math> . By Law of Cosines on <math>\triangle APB</math>, <math>\cos(\angle APB)=\frac{4\sqrt{65}}{65}</math> so <math>\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}</math>. Using <math>[\triangle ABC]=\frac{ab\sin(\angle C)}{2}</math> to find unknown areas, <math>[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}</math>. |
-Solution by Gart | -Solution by Gart |
Revision as of 14:24, 11 December 2021
Contents
Problem
Let be a convex quadrilateral with
,
, and
. Assume that the diagonals of
intersect at point
, and that the sum of the areas of triangles
and
equals the sum of the areas of triangles
and
. Find the area of quadrilateral
.
Diagram
Let and let
. Let
and let
.
Solution 1
Let and let
. Let
and let
. We easily get
and
.
We are given that , which we can now write as
Either
or
. The former would imply that
is a parallelogram, which it isn't; therefore we conclude
and
is the midpoint of
. Let
and
. Then
. On one hand, since
, we have
whereas, on the other hand, using cosine formula to get the length of
, we get
Eliminating
in the above two equations and solving for
we get
which finally yields
.
Solution 2
For reference, , so
is the longest of the four sides of
. Let
be the length of the altitude from
to
, and let
be the length of the altitude from
to
. Then, the triangle area equation becomes
What an important finding! Note that the opposite sides and
have equal length, and note that diagonal
bisects diagonal
. This is very similar to what happens if
were a parallelogram with
, so let's extend
to point
, such that
is a parallelogram. In other words,
and
Now, let's examine
. Since
, the triangle is isosceles, and
. Note that in parallelogram
,
and
are congruent, so
and thus
Define
, so
.
We use the Law of Cosines on and
:
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot
on
gives
and therefore
. Seeing that
, we conclude that
is a 3-4-5 right triangle, so
. Then, the area of
is
. Since
, points
and
are equidistant from
, so
and hence
-kgator
Just to be complete -- and
can actually be equal. In this case,
, but
must be equal to
. We get the same result. -Mathdummy.
Solution 3 (Another way to get the middle point)
So, let the area of triangles
,
,
,
. Suppose
and
, then it is easy to show that
Also, because
we will have
So
So
So
So
As a result,
Then, we have
Combine the condition
we can find out that
so
is the midpoint of
~Solution by (Frank FYC)
Solution 4 (With yet another way to get the middle point)
Denote by
. Then
.
Using the formula for the area of a triangle, we get
so
Hence
(note that
makes no difference here).
Now, assume that
,
, and
. Using the cosine rule for
and
, it is clear that
or
Likewise, using the cosine rule for triangles
and
,
It follows that
Since
,
which simplifies to
Plugging this back to equations
,
, and
, it can be solved that
. Then, the area of the quadrilateral is
--Solution by MicGu
Solution 5
As in all other solutions, we can first find that either or
, but it's an AIME problem, we can take
, and assume the other choice will lead to the same result (which is true).
From , we have
, and
, therefore,
By Law of Cosines,
Square
and
, and add them, to get
Solve,
,
-Mathdummy
Solution 6
Either or
. Let
. Applying Stewart's Theorem on
and
, dividing by
and rearranging,
Applying Stewart on
and
,
Substituting equations 1 and 2 into 3 and rearranging,
. By Law of Cosines on
,
so
. Using
to find unknown areas,
.
-Solution by Gart
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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